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Why does supersymmetry ensure that the entropy of a black hole calculated for g=0 can be extrapolated to be valid at finite string coupling?

+ 4 like - 0 dislike
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In ch. 22 of Zwiebach's first course in string theory, it is explained that a strategy to calculate the entropy of a black hole in string theory is to count the states for $g=0$ (no interactions) and then extrapolate the result to a regime of finite coupling. In ch.22.7 it is roughly outlined how it works for a 5d black hole in type IIB superstring theory with 5 dimensions compactified to circles, leading to the result

\(\frac{S_{bh}}{k} = 2 \pi \sqrt{NQ_1Q_5}\)

$Q_1$ is the charge generated by wrapping D1-branes around the compactified dimension $x^5$, $Q_5$ is generated by wrapping D5-branes around all 5 compactified dimensions, and $N$ is the momentum quantum number along $x^5$,

Supersymmetry is mentioned to be a prerequisite for the extrapolation to finite coupling regimes to be allowed.

Why is supersymmetry needed for this extrapolation to be valid, how can I see/understand this? Is the presence of supersymmetry necessary and sufficient?

asked Apr 21, 2014 in Theoretical Physics by Dilaton (4,295 points) [ revision history ]
edited Apr 21, 2014 by dimension10

*General comment:* Supersymmetry is very useful in field theory (or supergravity) because sufficient supersymmetry gives us powerful no-renormalization theorems, for example: zero vacuum energy. *Wrt OP:* I wonder if the role of SUSY is in providing BPS states which might be easy to count. Once counted, I don't see why one needs SUSY to claim that the number of states in the space is independent of the value of the coupling. It is, after all, the same Hilbert space. (Not authoritative. Just thinking out loud.)

Zwiebach's first course in string theory is for undergradutes, not graduates, making this an undergraduate question.

If you know enough physics, you can even ask on PhysicsOverflow on topic questions originally inspired by a popular TV show, for example about the mathematical technical details of an in the original source only hand-wavingly mentioned issue. So what ...?

@dilaton My misunderstanding, I thought the reader was expected to work this out for themselves from the material preceding this.

3 Answers

+ 5 like - 0 dislike

To provide a tad more explanation to what suresh1 says in his reply, since the mechanism here is as simple as it is fundamental to all things supersymmetric and index-theoretic:

It is a general fact that "the partition function of a supersymmetric system is a topological invariant". This was first discovered in the context of supersymmetric quantum mechanics, where the phenomenon is most clearly visible, but precisely the same general mechanism controls all supersymmetric quantum field theories and string theories, too.

Namely,consider a system with Hamiltonian \(H\) that "has a supercharge" hence for which there is an odd-graded self-adjoint operator \(D\) such that \(H = \tfrac{1}{2}\{D,D\} = D^2\)

Then consider the Euclideanized partition function of the system

\(Z_t = \mathrm{sTr}(\exp(-t H))\)

where we take the super-trace, hence the trace over even-graded states minus that over odd-graded states.

Now the point to notice is that all eigenstates of H of  non-vanishing eigenvalue appear in “supermultiplet” pairs of the same eigenvalue: if \(|\psi\rangle\) has eigenvalue E>0 under H, then

  1. \(D |\psi\rangle \neq 0\);

  2. also \(D |\psi\rangle\) has eignevalue E (since [H,D]=0).

Therefore all eigenstates for non-vanishing eigenvalues appear in pairs whose members have opposite sign under the supertrace. So only states with \(H |\psi\rangle = 0\) contribute to the supertrace. But if H and D are hermitean operators for a non-degenerate inner product, then it follows that \((H |\psi\rangle = 0) \Leftrightarrow (D|\psi\rangle = 0)\) and hence these are precisely the states which are also annihilated by the supercharge (are in the kernel of D), hence are precisely only the supersymmetric states.

On these now the weight \(\exp(-t D^2) = 1\) and hence the supertrace over this “Euclidean propagator” simply counts the number of supersymmetric states, signed by their fermion number. 

Notice that this result is "topological" in that it does not actually depend on the geometric background encoded by the Hamiltonian (it's energy spectrum and hence the couplings which this encodes) In particular it does not depend on the worldline length t itself.

It seems that this simple argument goes back at least all the way to 

  • H. MacKean, Isadore SingerCurvature and eigenvalues of the Laplacian, J. Diff. Geom. 1 (1967)

It became popular with the advent of supersymmetric quantum mechanics due to

  • Luis Alvarez-GauméSupersymmetry and the Atiyah-Singer index theorem, Comm. Math. Phys. Volume 90, Number 2 (1983), 161-173. (Euclid)

In the mathematics literature this was then picked up in 

  • Ezra GetzlerPseudodifferential operators on supermanifolds and the Atiyah-Singer index theorem, Comm. Math. Phys. 92 (1983), 163-178. (pdf)

and so today a standard textbook account for this is

(see around prop. 3.48, 3-50). Mathematicians speak of the "heat kernel technique for index theory" or the like. For more see on the nLab at index.

Now to come to the string: superstring propagation may be thought of as essentially being supersymmetric quantum mechanics on loop space (this is how Witten was brought to supersymmetric QM in the first place). The Dirac operator now is the Dirac-Ramond operator. The partition function of the superstring is hence formally directly analogous, just richer, to the above. In particular if a sector has a supercharge, then only the supersymmetric ground states contribute to the partition function.

In particular for the type II superstring in the NS-R sector or else for the heterotic string, of the two possible worldsheet Dirac-Ramond supercharges only one exists (the right-moving one, say). Hence the partition function of the superstring in this case -- called the Witten genus -- may be thought of as being a generating functional for counting of one-half supersymmetric string states, by precisely the above kind of argument.

But one-half supersymmetric string states have an important interpretation in terms of the effective target space supergravity theory: theory are the BPS-states. Hence the partition function of the one-half-supersymmetric superstring (type II in NS-R sector else heterotic) counts BPS-states in target space.

Lecture notes expanding on this include for instance

This is finally where the black hole entropy comes in: suitably extremal black holes in supergravity (= strongly coupled strings) correspond indeed to BPS states in the weakly coupled theory (certain half-supersymmetric D-brane configurations). The former are counted by the Witten genus. But since -- by the above argument -- that does not actually depend on "geometric" data such as the  coupling constant, one knows that it remains the same even in the strong coupling regime in which there is a black hole.

This is what one means when one says that BPS states are "protected" from changes of the coupling constant.

More pointers are on the nLab at black holes in string theory

answered Apr 22, 2014 by Urs Schreiber (5,785 points) [ no revision ]

Thanks Urs for this immensely nice and rich answer, it gives me quite something to dig into :-)

+1 Adds a lot more to my (rather bland) answer. 

+ 4 like - 0 dislike

The protected objects are supersymmetric indices (the analog of the Witten index). These are signed countings and take the form $(n_b -n_f)$ where I use a subscript $b$ to indicate a "bosonic" contribution counted with a positive sign and the subscript $f$ to indicate a "fermionic" contribution. So in situations where $n_f$ is zero, the counting by the index is equal to the unsigned counting. It might suffice to have a situation where $n_b\gg n_f$. In such situations, one expects to be able to reproduce the entropy of a zero-temperature black hole using the index as Strominger and Vafa did in the example that you mention. There are indeed situations where the index does not reproduce the entropy.

The supersymmetric index can be shown to be independent of, say the string coupling. Then one can extrapolate answers computed at weak string coupling (where one is counting D-brane configurations) to strong string coupling where one might be dealing with black holes.

Note added: The index computed depends on the precise context. To illustrate this: Consider 4D compactifications of string theory with N=4 susy -- there are different indices (elliptic genus, helicity supertraces,...) that count $1/2$ and $1/4$ BPS states in this theory.  This review by Ashoke Sen will provide more details. This system also provides examples where the index counting matches entropy (when the model has a heterotic dual) and the index counting doesn't match (when the model does not have a heterotic dual). Refined countings incorporating discrete symmetries have lead to amazing connections with the sporadic finite groups Mathieu group $M_{24}$ and the Conway group $Co_0$.

answered Apr 22, 2014 by suresh (1,535 points) [ revision history ]
edited Apr 23, 2014 by suresh
+ 3 like - 0 dislike

Urs Schreiber and suresh1 have explained the general answer to the question. As mentionned by suresh1, the precise index to compute depends on the precise context. I simply give the answer for the precise case mentionned in the question, i.e. IIB on T^5 with a D1-D5 system. In the 5 non-compact dimensions, we have a maximally supersymmetric supergravity (32 real supercharges). The black hole obtained from the D1-D5-N system is 1/8 BPS. This system can be described at low energy by a 2d N=(4,4) SCFT and the relevant index is a modified elliptic genus (the standard elliptic genus defined for a 2d N=(2,2) SCFT vanishes for N=(4,4) due do the extrasymmetry but this extrasymmetry also guarantees that some modification of the elliptic genus, which is not an index in a general N=(2,2) SCFT, becomes an index in the N=(4,4) case). This story is very well explained by Maldacena, Moore and Strominger in http://arxiv.org/abs/hep-th/9903163 .

answered Apr 24, 2014 by 40227 (4,660 points) [ revision history ]
edited Apr 25, 2014 by dimension10

I have corrected a link in your answer (the full stop), if that's ok with you.   

Thanks for the correction of the link.

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